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Let $p:X\rightarrow Y$ be a closed continuous surjective map such that $p^{-1}(y)$ is compact for each $y\in Y$. Show that if $X$ is second countable then $Y$ is second countable.

Let $y\in Y$ and $U$ be an open set in $Y$. Let $x\in X$ such that $p(x)=y$.

As $p$ is continuous, $p^{-1}(U)$ is open and contains $x$.

As $X$ is second countable there exists a countable collection of open sets $\{U_n\}$ being basis for $X$.

We have $x\in X$ and $x\in p^{-1}(U)$. So, there exists $n\in \mathbb{N}$ such that $x\in U_n\subset p^{-1}(U)$..

Do the same for all elements of $p^{-1}(y)$. We then have $p^{-1}(y)\subset \bigcup_{n=1}^{\infty} U_n$.

As $p^{-1}(y)$ is compact there exists a finite subcover. We then have $p^{-1}(y)\subset \bigcup_{i=1}^n U_i\subset p^{-1}(U)$.

Set $M=\bigcup_{i=1}^n U_i$ for simplicity of notation.

Claim is that $y\in p(M^c)^c\subset U$.

Suppose $y\notin p(M^c)^c$ then $y\in p(M^c)$ then $y=p(a)$ for some $a\in M^c$. But then $a\in p^{-1}(y)\subset M$. So, $a\in M$ and $a\in M^c$ a contradiction. So, we do have $y\in p(M^c)^c$

Suppose $m\in U^c$ and $p(a)=m$. We then have $a\in p^{-1}(m)\subset p^{-1}(U^c)\subset p^{-1}(U)^c\subset M^c$. So, $p(a)=m$ for some $a\in M^c$ i.e., $m=p(a)\in p(M^c)$ and so $y\notin p(M^c)^c$.

So, $y\in p(M^c)^c\subset U$.

So, we have a countable collection $$\left\{p\left(\bigcap_{i=1}^n U_i^c\right)^c:n\in \mathbb{N}\right\}$$ is a countable basis for $Y$.

Let me know if there are any gaps.

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The basic idea is fine, but there are some problems with the notation, and the organization could be better.

The worst notational problem is when you write that $p^{-1}(y)\subset\bigcup_{n=1}^\infty U_n$. Since the sets $U_n$ in this union were obtained as nbhds of points of $p^{-1}[\{y\}]$, you seem to be assuming here that $p^{-1}[\{y\}]$ is countably infinite, when in fact it could be finite or uncountable. Moreover, your countable base for $X$ was $\{U_n:n\in\Bbb N\}$, so the union of the sets $U_n$ is all of $X$; this does indeed contain $p^{-1}[\{y\}]$, but in general it will not be a subset of $p^{-1}[U]$. To avoid this problem you could say that for each $x\in p^{-1}[\{y\}]$ there is an $n(x)\in\Bbb N$ such that $x\in U_{n(x)}\subseteq p^{-1}[\{y\}]$. Then you could observe that $\{U_{n(x)}:p(x)=y\}$, being an open cover of the compact set $p^{-1}[\{y\}]$, has a finite subcover $\{U_{n(x_1)},\ldots,U_{n(x_m)}\}$ for some $x_1,\ldots,x_m\in p^{-1}[\{y\}]$, and let $M=\bigcup_{k=1}^mU_{n(x_k)}$.

I would change the organization completely: I would start with a countable base $\mathscr{B}$ for $X$. I would then show that if we take the closure of $\mathscr{B}$ under finite unions, we still have a countable base for $X$, so we might as well assume that $\mathscr{B}$ is already closed under finite unions. Then I would prove that

$$\big\{Y\setminus p[X\setminus B]:B\in\mathscr{B}\big\}$$

is a countable base for $Y$.

That proof is essentially your argument. Let $U$ be a non-empty open set in $Y$, and let $y\in U$. For each $x\in p^{-1}[\{y\}]$ there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq p^{-1}[U]$. The family $\{B_x:p(x)=y\}$ is an open cover of the compact set $p^{-1}[\{y\}]$, so there is a finite $F\subseteq p^{-1}[\{y\}]$ such that $p^{-1}[\{y\}]\subseteq\bigcup\{B_x:x\in F\}$. And $\mathscr{B}$ is closed under finite intersections, so $\bigcup\{B_x:x\in F\}\in\mathscr{B}$; let $B(y)=\bigcup\{B_x:x\in F\}$. (This $B(y)$ corresponds to your $M$.)

$X\setminus B(y)$ is closed in $X$, so $p[X\setminus B(y)]$ is closed in $Y$, and $Y\setminus p[X\setminus B(y)]$ is open in $Y$. Moreover, $p^{-1}[\{y\}]\cap\big(X\setminus B(y)\big)=\varnothing$, so $y\notin p[X\setminus B(y)]$, and therefore $Y\setminus p[X\setminus B(y)]$ is an open nbhd of $y$. All that remains is to show that $Y\setminus p[X\setminus B(y)]\subseteq U$, which can be done directly. Let $u\in Y\setminus p[X\setminus B(y)]$; there is some $x\in X$ such that $u=p(x)$. Clearly $x\notin X\setminus B(y)$, so $x\in B(y)\subseteq p^{-1}[U]$, and $u=p(x)\in U$.

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